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PostPosted: May 09, 2007 9:52 am 
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I saw this recently and thought it had some very interesting implications:

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Promising Pyrochemical Actinide/Lanthanide Separation Processes Using Aluminum

Volume 153 · Number 3 · July 2006 · Pages 253-261
Technical Paper

Olivier Conocar, Nicolas Douyere, Jean-Paul Glatz, Jérôme Lacquement, Rikard Malmbeck, Jérôme Serp

Thermodynamic calculations have shown that aluminum is the most promising metallic solvent or support for the separation of actinides (An) from lanthanides (Ln). In molten fluoride salt, the technique of reductive extraction is under development in which the separation is based on different distributions of An and Ln between the salt and metallic Al phases. In this process molten aluminum alloy acts as both a reductant and a solvent into which the actinides are selectively extracted. It was demonstrated that a one-stage reductive extraction process, using a concentrated solution, allows a recovery of more than 99.3% of Pu and Am. In addition excellent separation factors between Pu and Ln well above 10^3 were obtained. In molten chloride media similar separations are developed by constant current electrorefining between a metallic alloy fuel (U60Pu20-Zr10Am2Nd3.5Y0.5Ce0.5Gd0.5) and an Al solid cathode. In a series of demonstration experiments, almost 25 g of metallic fuel was reprocessed and actinides collected as An-Al alloys on the cathode. Analysis of the An-Al deposits confirmed that an excellent An/Ln separation (An/Ln mass ratio = 2400) had been obtained. These results show that Al is a very promising material to be used in pyrochemical reprocessing of actinides.


If this article is correct, and aluminum can be used in a reductive extraction process to remove plutonium and americium from a fluoride salt mixture, then this could have some very interesting implications.

One of my first thoughts was that conventional uranium oxide spent fuel could be reprocessed rather simply as a fluoride rather than as an oxide. Recall that all uranium oxide is converted to a fluoride in order to be enriched, and then converted back to an oxide to serve as fuel.

The process I can imagine might work something like this:

  1. Spent nuclear fuel would be fluorinated to convert oxides of uranium, plutonium, other TRUs, and fission products into fluorides.
  2. Plutonium, americium, and other TRUs would be extracted as metals using aluminum as the reductive extractor. Basically, metallic aluminum would become aluminum fluoride, and Pu and Am would turn from fluorides into metals.
  3. Remaining uranium would be extracted by fluorination from uranium tetrafluoride (UF4) to uranium hexafluoride (UF6). Neptunium, molybdenum, and technetium would also come out of the solution as volatile hexafluorides. The isotopic composition of the uranium at this point would be closer to natural uranium (~1% U-235) than its original enrichment (~3% U-235).
  4. Remaining fission product fluorides, comprising about 3% of the original volume of the spent fuel, would be separated from one another by distillation. Useful stable fission products (like rhodium) could be sold, and those that might have value as radioisotopes (Cs and Sr) might be sold as well.

The extracted metallic TRUs could then be chlorinated and destroyed in a chloride fast reactor.

Does anyone have more information on this aluminum reductive extraction process? It seems somewhat similar to the old ORNL idea of using bismuth to reductively extract protactinium from the salt of a fluoride reactor.


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PostPosted: Sep 12, 2007 9:34 pm 
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Another thread is talking about removing plutonium, and I wanted to point out this very promising technique.


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PostPosted: Jan 01, 2011 8:55 pm 
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Purdue researchers demonstrate their method for producing hydrogen by adding water to an alloy of aluminum and gallium. Here is a video demonstrating the process.
Quote:
This reaction splits the oxygen and hydrogen contained in water, releasing hydrogen in the process.

The gallium is critical to the process because it hinders the formation of a skin normally created on aluminum's surface after oxidation. This skin usually prevents oxygen from reacting with aluminum, acting as a barrier. Preventing the skin's formation allows the reaction to continue until all of the aluminum is used.


In the electrorefining of metallic alloy fuel onto solid Al cathodes, the same type of skin formation gum up occurs reducing the efficiency of the electrorefining process as follows:

Quote:
In such an electrolytic process, the rate of the alloy formation depends on the
diffusion of the involved elements in and through the solid alloy phase. Thus
with increasing thickness of the alloy, further deposition of An as An-Al alloy
becomes more and more difficult. If the deposition rate of An is too high, a
depletion of Al occurs at the surface and the actinides will deposit as pure metals,
at potentials shifted towards more negative values. This must be avoided, since
for an efficient grouped separation of actinides from lanthanides a potential more
anodic than -1.25 V vs. Ag/AgCl is required. The applied current has to be
lowered as soon as the cathodic potential becomes too negative. However, a
point is finally reached when it is no more possible to maintain the cathodic
potential positive enough to ensure the separation of An from Ln. This limits the
amount of An which can be recovered onto Al.



Image
Closed cell aluminium foam with a small cell size

If the cathodes are formed from aluminum-gallium alloy metal foam where the total surface area of the cathode is huge and unobstructed by skin formation, I speculate that the performance and efficiency of this electrorefining process will be spectacular.

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PostPosted: Jan 02, 2011 4:55 am 
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That sounds like a sensible strategy. Perhaps discuss some small changes?

1. 4HF+AnO2= AnF4+2H2O. Now you have tetrafluorides and can already vacuum distill the UF4, maybe also grab Np as NpF4. The volatile fission products like iodine will also be readily seperated in cold and carbon traps. This should leave the Pu, Am, and Cm largely as trifluorides.
2. Al+AnF3=AlF3+An. Perfect reaction fit! And the free energy of AlF3 is nicely between the actinides and the lanthanides, too.
3. Is flame fluorination really necessary now? You got almost all actinides already. If we can't vacuum distill NpF4 then fluorination will be a good way to grab it now.
4. Remaining fission product stuff to be distilled? Why do the distillation at the end? You'll need quite high temperatures to seperate the lanthanide fluorides. Further reduction with magnesium followed by zone refining and/or metal distillation is also an option.


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